identifying compounds How could I tell this is actually sodium chloride?

Elements and colour When different elements are heated strongly the atoms within them absorb energy They then emit light of a characteristic frequency

Identifying elements If the frequency of light emitted is in the visible region of the spectrum we will see the colour emitted and can identify the element.

Identifying elements Take a wire and dip into metal salt and then touch the blue part of the Bunsen flame with the compound. The wire should be dipped in concentrated HCl to clean it

Complete your table Metal Cation Colour Sodium Orange Calcium Red Copper Blue Barium Green Potassium Lilac Lithium Crimson Magnesium no colour Aluminium Iron (II) or iron (III)

Other tests for cations Some metals do not have a distinctive colour. We can use other analytical methods. Adding sodium hydroxide can produce precipitates (solids) with distinctive colours.

Precipitates Metal ion Colour of precipitate Other features Aluminium white redissolves in excess hydroxide Calcium also gives a red flame in Bunsen burner Magnesium   Copper (II) Cu2+ blue Iron (II) Fe2+ pale green Iron (III) Fe3+ brown

Anions The anion is the rest of the ionic compound once you have identified the cation. Common anions include Sulfate SO42- Carbonate CO32- Chloride Cl- Bromide Br- Iodide I-

Summary of anion tests Anion Test Result Carbonate Add hydrochloric acid Solution fizzes (produces some bubbles) Sulfate Add barium chloride White precipitate formed Chloride Add silver nitrate Bromide Cream precipitate formed Iodide Yellow precipitate formed

Putting the two together Once we have separately identified the anion and cation we can write the name of the compound. For example a compound that gives a lilac flame test and a white precipitate on addition of silver nitrate is… POTASSIUM CHLORIDE!

What am I? The cation gives no precipitate and bright orange flame The anion gives white precipitate with silver nitrate

What am I? The cation gives a lilac flame and no precipitate The anion gives a yellow precipitate with silver nitrate

What am I? The flame tests give no colour. The precipitate with sodium hydroxide is orange/brown The anion gives a white precipitate when barium chloride is added

What am I? The flame test gives no colour. There is a white precipitate with sodium hydroxide that redissolves in excess hydroxide The anion gives a cream precipitate with silver nitrate

A white precipitate was formed. Dilute sodium hydroxide solution was added to the sample (ammonia solution would also be appropriate) A white precipitate was formed. The precipitate dissolved when excess sodium hydroxide solution was added to give a colourless solution (it would not have dissolved if excess ammonia solution was added) I.G.C.S.E. only Aqueous Cation

Aluminium

A white precipitate formed. A small amount of dilute sodium hydroxide solution was added to the sample. A white precipitate formed. When excess sodium hydroxide solution was added the precipitate did not dissolve. When tested with dilute ammonia solution no/very little precipitate was formed. I.G.C.S.E. only Aqueous Cation

Magnesium

The gas was found to turn damp red litmus paper blue. A small amount of dilute sodium hydroxide or ammonia solution was added to the sample. The solution was heated gently and gave off a gas. The gas was found to turn damp red litmus paper blue. Aqueous Cation

Ammonium – gas is ammonia

A pale blue precipitate formed. A small amount of dilute sodium hydroxide or ammonia solution was added to the sample. A pale blue precipitate formed. When excess ammonia solution was added the precipitate dissolved and a deep blue solution was formed. Excess sodium hydroxide solution had no effect. Aqueous Cation

Copper

A green precipitate is formed. Dilute sodium hydroxide solution (or ammonia solution) is added to the sample. A green precipitate is formed. Aqueous Cation

Iron(11)

A red-brown precipitate is formed. Dilute sodium hydroxide solution (or ammonia solution) is added to the sample. A red-brown precipitate is formed. Aqueous Cation

Iron(111)

A pale yellow precipitate formed. An equal amount of dilute nitric acid was added to a small sample of the solution. Then a few drops of silver nitrate solution were added. A pale yellow precipitate formed. I.G.C.S.E. only ANION

iodide

A white precipitate formed. An equal amount of dilute nitric acid was added to a small sample of the solution. Then a few drops of silver nitrate solution were added. A white precipitate formed. ANION

Chloride

A white precipitate was formed. An equal amount of dilute hydrochloric acid was added to a small sample of the solution. Then a few drops of barium nitrate were then added. A white precipitate was formed. ANION

sulfate

A little dilute hydrochloric acid was added to the solid/solution. The mixture bubbled. The gas was blown thorough lime water and it turned milky. ANION

Carbonate

GAS This is a colourless gas with a sharp smell. A sample was collected in a test tube. Damp red litmus paper was inserted into the sample. The gas turned damp red litmus paper blue. GAS

ammonia

GAS This is a colourless gas. A sample was bubbled through limewater (which is a clear liquid). The gas reacted with the limewater and turned it cloudy/milky. GAS

Carbon dioxide

The damp litmus paper then bleached to a white colour. This is a greenish gas. A sample was collected in a test tube, in the fume cupboard. Damp litmus paper was inserted into the sample. The damp litmus paper then bleached to a white colour. GAS

chlorine

GAS This is a colourless gas. A sample was collected an upturned test tube and a lighted splint was held underneath. The gas burnt with a loud squeaky pop. GAS

hydrogen

GAS This is a colourless gas. A sample was collected an test tube and a glowing splint was inserted. The splint burst into flames. GAS

oxygen

What ions you need to be able to identify Cations Anions Sodium Potassium Lithium Barium Calcium Iron(11) Iron (111) Copper (11) Ammonium Sulphate Carbonate Chloride Bromide Iodide

Half Equations LO - To be able to write balanced equations to show what is happening electronically at the positive and negative electrode. What is the charge for a) chlorine and b) calcium in the following compound? CaCl2

The half equation for the formation of water is Half equations For the reaction NaOH + HCl → NaCl + H2O The half equation for the formation of water is H+ + -OH → H2O The half equation for the formation of this salt is Na+ + Cl- → NaCl NOTE – Ions are on the left and they produce a neutral compound shown to the right.

WE REPRESENT WHAT IS HAPPENING AT EACH ELECTRODE USING HALF EQUATIONS The oxidation of iodine at the positive electrode 2I- → I2 + 2e

Half Equations for Negative Ions The ion is written to the left side of the equation arrow e.g. I- → Usually negative ions form gases so we write the product as a diatomic molecule and balance the equation in terms of atoms e.g. 2I- → I2 Finally balance the equation in terms of charge by adding electrons e.g. 2I- → I2 + 2e-

Half Equations for Positive Ions The ion is written to the left side of the equation arrow e.g. Ca2+ → Metal ions lose their charge and become pure metals not diatomic molecules e.g. Ca2+ → Ca Finally balance the equation in terms of charge by adding electrons e.g. Ca2+ + 2e- → Ca

Questions Write balanced half equations for the following ions at the cathode. Pb2+, potassium, hydrogen, aluminium Write balanced half equations for the following ions at the anode. chlorine, bromine, oxygen EXTENSION Write half equations for the electrolysis of TiF4 → Ti + 2F2

Spot the mistake…….

Sodium bromate is a compound of sodium, bromine and oxygen. A sample of sodium bromate contains 2.3 g of sodium, 8.0 g of bromine and 4.8 g of oxygen. Calculate the empirical formula of sodium bromate.

What happens at the cathode when you electrolyse NaCl solution? Why can’t you do electrolysis on solid ionic compounds? The anode is...? What group is sodium? What happens at the cathode when you electrolyse NaCl solution? Na metal plus water makes what? What is an electrode? What is an electrolyte? Positive attracts? Which end do electrons come out of a cell? Why do hydrogen form diatomic molecules?

complete Products formed Electrode PbBr2 HCl NaCl CuSO4 Cathode Anode

The Faraday A Faraday is one mole of electrons, and is equivalent to 96 500C (Coulombs) A current of 1A = 1C per second flowing For example, Cu2+ + 2e  Cu 1 mole of Cu2+ ions reacts with 2 Faradays of electrons, to produce 1 mole of Cu

Quantity of electricity in coulombs = current in amps x time in seconds Q (C) = I (A) x t (s)

Example How much copper is deposited if a current of 0.2 Amps is passed for 2 hours through a copper(II) sulphate solution ?

current of 0.2 Amps is passed for 2 hours At the cathode: Cu2+(aq) + 2e-  Cu(s) Q = I x t = 0.2 x (2 x 60 x 60) =1440 Coulombs 1 mole electrons = 96500 Coulombs So, 1440 / 96500 = 0.0149 moles of electrons (Faradays)

Example moles of electrons passed through circuit = 0.01492 Cu2+(aq) + 2e-  Cu(s) From equation, it takes two moles of electrons to form one mole of copper moles copper = 0.01492 / 2 = 0.00746

Example moles Cu = 0.00746 mass of Cu = moles x Ar = 0.00746 x 64 = 0.4775g of Cu deposited.

‘How To” Guide Write out relevant half equation Work out coulombs of electrons flowing (Q = It) Convert C into moles of electrons (Faradays) (Q/96500) Work out moles of product using ratio from equation Convert into mass (mass = moles x Ar)

In an electrolysis of sodium chloride solution experiment a current of 2 A was passed for 2 minutes. (a) Calculate the volume of chlorine gas produced. (b) What volume of hydrogen would be formed? (c) In practice the measured volume of chlorine can be less than the theoretical value. Why?

(a) Calculate the volume of chlorine gas produced. Electrode equations: (-) cathode 2H+ + 2e-  H2 (+) anode 2Cl-  Cl2 + 2e  (a) Calculate the volume of chlorine gas produced. Q = I x t, so Q = 2 x 2 x 60 = 240 C 240 C = 240 / 96500 = 0.002487 mol electrons this will produce 0.002487 / 2 = 0.001244 mol Cl2 (two electrons/molecule) vol = mol x molar volume = 0.001244 x 24000 = 29.8 cm3 of Cl2 

(b) What volume of hydrogen would be formed? 29.8 cm3 of H2 because two electrons transferred per molecule, same as chlorine. (c) In practice the measured volume of chlorine can be less than the theoretical value. Why? chlorine is moderately soluble in water and also reacts with the sodium hydroxide formed.

In the electrolysis of molten sodium chloride 60 cm3 of chlorine was produced. Calculate ... (a) how many moles of were chlorine produced? (b) what mass of sodium would be formed? (c) for how long would a current of 3 A in the electrolysis circuit have to flow to produce the 60cm3 of chlorine?

(a) how many moles of chlorine produced? 60 / 24000 = 0.0025 mol Cl2 

(b) what mass of sodium would be formed? from the electrode equations 2 mol sodium will be made for every mole of chlorine so 0.0025 x 2 = 0.005 mol sodium will be formed. Ar(Na) = 23 mass = mol x atomic or formula mass = 0.005 x 23 = 0.115g Na

(c) for how long would a current of 3 A in the electrolysis circuit have to flow to produce the 60cm3 of chlorine? To produce 0.0025 mol of Cl2 you need 0.005 mol of electrons 0.005 mol electrons = 0.005 x 96500 coulombs = 482.5 C Q = I x t, so 482.5 = 2 x t, therefore t = 482.5 / 3 = 161 s (to nearest second)

i.e. 0.0448/2 moles = 0.0224 moles Calculations for you to try. Calculate the mass of hydrogen formed when a current of 0.4 A is passed through hydrochloric acid solution for 3 hours. Step 1: Q = I x t = 0.4 x 3 x 60 x 60 = 4320 C Step 2: calculate moles of electrons used – 4320/96500 = 0.0448 Step 3: Use the ionic half equation to calculate the moles of hydrogen released 2H+(aq) + 2e-  H2(g) moles of electrons = moles of hydrogen released/2 i.e. 0.0448/2 moles = 0.0224 moles Step 4: Use the number of moles to calculate the mass of hydrogen released MOLES = MASS (g) / Mr of H2 0.0224 = MASS (g) / 2 0.0224 x 2 = MASS (g) MASS of hydrogen = 0.0448 g

Example 2: Calculate the volume of chorine produced when a current of 2 A is passed though sodium chloride solution for 5 hours 40 minutes. The molar volume of a gas (at RTP) is 24 000 cm3. Step 1: Q = I x t = 408 000 C Step 2: calculate moles of electrons used – 408 000/96500 = 4.23 moles Step 3: Use the ionic half equation to calculate the moles of chlorine released 2Cl-(aq) - 2e-  Cl2(g) moles of electrons = moles of chlorine released/2 i.e. 2.12 moles i.e. 2.12 moles Step 4: Use the number of moles to calculate the volume of chlorine released MOLES = volume (cm ) / 24 000 3 2.12 = volume (cm ) / 24 000 3 2.12 x 24 000 = volume (cm ) 3 Volume of chlorine = 50760 cm 3